Examine the character of the rest point (0,0,0) of the…

Examine the character of the rest point $(0,0,0)$ of the system $$\dot{x}=x-3y+4z,\qquad\dot{y}=4x-7y+8z,\qquad\dot{z}=6x-7y-7z.$$

Step 1 · Coefficient matrix

$$A=\begin{pmatrix}1&-3&4\\4&-7&8\\6&-7&-7\end{pmatrix}.$$

Step 2 · Characteristic equation

Expanding $\det(A-\lambda I)=0$ gives $$\lambda^3+13\lambda^2+79\lambda+67=0.$$

Step 3 · Solving the characteristic equation

Testing $\lambda=-1$: $-1+13-79+67=0$. Factoring, $\lambda^3+13\lambda^2+79\lambda+67=(\lambda+1)(\lambda^2+12\lambda+67)$, and the quadratic gives $\lambda=-6\pm\sqrt{31}\,i$. The three eigenvalues are $$\lambda_1=-1,\qquad\lambda_{2,3}=-6\pm\sqrt{31}\,i.$$

Step 4 · Classification of the rest point

All eigenvalues have negative real parts: $\operatorname{Re}(\lambda_1)=-1<0$, $\operatorname{Re}(\lambda_{2,3})=-6<0$. Since one eigenvalue is real and two are complex conjugates with non-zero imaginary part, the origin is a stable focus-node (stable spiral node): trajectories spiral toward the origin.

Step 5 · Eigenvector for $\lambda_1=-1$

Solving $(A+I)\mathbf{v}=0$ (row-reducing $\begin{pmatrix}2&-3&4\\4&-6&8\\6&-7&-6\end{pmatrix}$) gives $v_2=9v_3$ and $v_1=\tfrac{23}{2}v_3$. Taking $v_3=2$: $$\mathbf{v}^{(1)}=\begin{pmatrix}23\\18\\2\end{pmatrix}.$$

Step 6 · Eigenvector for $\lambda_2=-6+\sqrt{31</h2> <p>\,i$} Solving the corresponding complex eigenvector equation (using rows 2 and 3 with $v_1=1$) gives $$\mathbf{v}^{(2)}=\begin{pmatrix}1\\2\\\tfrac{-1+\sqrt{31}\,i}{4}\end{pmatrix} =\underbrace{\begin{pmatrix}1\\2\\-\tfrac{1}{4}\end{pmatrix}}_{\mathbf{p}} +i\underbrace{\begin{pmatrix}0\\0\\\tfrac{\sqrt{31}}{4}\end{pmatrix}}_{\mathbf{q}}.$$

Step 7 · General real solution

With $\alpha=-6$, $\beta=\sqrt{31}$: $$x(t)=23C_1 e^{-t}+e^{-6t}(C_2\cos\sqrt{31}\,t+C_3\sin\sqrt{31}\,t),$$ $$y(t)=18C_1 e^{-t}+2e^{-6t}(C_2\cos\sqrt{31}\,t+C_3\sin\sqrt{31}\,t),$$ $$z(t)=2C_1 e^{-t}+e^{-6t}\!\left[ \left(-\tfrac{C_2}{4}+\tfrac{\sqrt{31}}{4}C_3\right)\cos\sqrt{31}\,t +\left(-\tfrac{C_3}{4}-\tfrac{\sqrt{31}}{4}C_2\right)\sin\sqrt{31}\,t \right].$$

Step 8 · Behaviour of trajectories

• The real mode $e^{-t}\mathbf{v}^{(1)}$ decays at rate 1.
• The complex modes decay at the faster rate $e^{-6t}$ while oscillating with angular frequency $\omega=\sqrt{31}$ and period $T=\tfrac{2\pi}{\sqrt{31}}$.
• As $t\to+\infty$, both modes tend to zero, so all trajectories converge to the origin; as $t\to-\infty$ they diverge.
• The dominant behaviour for large $t$ (slowest decay) is governed by $e^{-t}$ along $\mathbf{v}^{(1)}=(23,18,2)^\top$; trajectories generically approach the origin tangentially to this eigendirection.

Step 9 · Stability analysis

Since $\operatorname{Re}(\lambda_k)<0$ for all $k=1,2,3$, the rest point is stable in the sense of Lyapunov and asymptotically stable: every trajectory converges to the origin as $t\to+\infty$.

Step 10 · Conclusion

$$\boxed{ (0,0,0)\text{ is a \textbf{stable focus-node} (asymptotically stable).} }$$ The eigenvalues are $\lambda_1=-1$ and $\lambda_{2,3}=-6\pm\sqrt{31}\,i$, all with negative real parts. The real eigenvalue contributes a non-oscillatory decaying mode along $\mathbf{v}^{(1)}=(23,18,2)^\top$, while the complex pair contributes an oscillatory mode decaying at the faster rate $e^{-6t}$ with angular frequency $\omega=\sqrt{31}$. All trajectories converge to the origin, approaching it tangentially to the slowest eigendirection.

Summary Table

Additional Notes

• In three dimensions the equilibrium is classified by the mix of eigenvalue signs (or, for a repeated eigenvalue, by its Jordan structure), which determines the dimensions of the stable and unstable manifolds.